Problem 22 - Elementary Algebra

Question

For the following problems, solve the equations. $$ \sqrt{3 m-5}=\sqrt{2 m+1} $$

Step-by-Step Solution

Answer

Answer: m = 6

Step 1: Rewrite the equation

Rewrite the given equation, so we can easily work with it. $$ \sqrt{3 m-5}=\sqrt{2 m+1} $$

Step 2: Square both sides

To eliminate the square roots, we can square both sides of the equation. It's crucial to remember that squaring both sides should be done cautiously since it can sometimes introduce extraneous solutions. $$ (\sqrt{3 m-5})^2=(\sqrt{2 m+1})^2 $$

Step 3: Simplify the equation

After squaring both sides, the square roots will cancel out, leaving us with a simpler equation. $$ 3 m-5=2 m+1 $$

Step 4: Solve for m

We can now isolate the variable m by moving all the m terms to one side and the constants to the other side. $$ 3m - 2m = 1 + 5 $$ $$ m = 6 $$

Step 5: Check the solution

It is essential to check our solution by plugging it back into the original equation, as squaring both sides can sometimes introduce extraneous solutions. Put m=6 into the original equation: $$ \sqrt{3(6)-5}=\sqrt{2(6)+1} $$ $$ \sqrt{13}=\sqrt{13} $$ Our solution is correct since the left side of the equation is equal to the right side. Therefore, the solution is m = 6.

Key Concepts

Squaring Both Sides

Solving radical equations often involves the simple yet powerful step of squaring both sides. When faced with an equation involving square roots, like \( \sqrt{3m-5} = \sqrt{2m+1} \), we use squaring to remove the radicals.
The idea is straightforward:

• Square the expression on the left-hand side: \((\sqrt{3m-5})^2\).
• Square the expression on the right-hand side: \((\sqrt{2m+1})^2\).

After squaring, the radicals disappear, changing our equation to one that is easier to work with:\[3m - 5 = 2m + 1\]However, it's important to be careful. Sometimes, squaring can introduce extra, or extraneous, solutions. These need to be checked later. For now, we've gone from a tricky radical equation to a simple linear one.

Checking Extraneous Solutions

Extraneous solutions are false solutions that come from the process of solving the equation, often due to squaring both sides. Even if the calculations seem correct, it's crucial to verify that these solutions satisfy the original equation. This is because squaring can sometimes make the equation valid for more values than the original form.
Let's consider our example result: \( m = 6 \). To check this solution:

• Substitute \( m = 6 \) back into the original equation: \( \sqrt{3(6) - 5} = \sqrt{2(6) + 1} \).
• Simplify both sides: \( \sqrt{13} = \sqrt{13} \).

The equation holds true, confirming that \( m = 6 \) is indeed a valid solution. Checking potential solutions ensures that we don't erroneously accept false ones introduced by our operations.

Simplifying Equations

Once the equation has been squared, the next step involves simplification. This transformation is key to making the problem manageable and moving toward the solution.After eliminating the square roots by squaring, you are left with:\[3m - 5 = 2m + 1\]The process involves:

• Gathering all terms involving the variable on one side and constants on the other:\(3m - 5 = 2m + 1\).
• Rearranging gives us \( 3m - 2m = 1 + 5 \).
• Solving the resulting equation: \( m = 6 \).

This simplification transforms the initial complex equation into a simple form, allowing us to find \( m \). Having mastered these steps, you can confidently handle similar radical equations in the future.